Dimension of Bohr Radius
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Theorem
The Bohr radius has the dimension $\mathsf L$.
Proof
By definition, the Bohr radius is:
- $a_0 = \dfrac {\varepsilon_0 h^2} {\pi \E^2 m_\E}$
where:
- $\varepsilon_0$ denotes the vacuum permittivity
- $h$ denotes Planck's constant
- $\E$ denotes the elementary charge
- $m_\E$ denotes the electron rest mass
We have:
\(\ds \varepsilon_0\) | \(\text {has dimension}\) | \(\ds \mathsf {M^{-1} L^{-3} T^4 I^2}\) | Definition of Vacuum Permittivity | |||||||||||
\(\ds h\) | \(\text {has dimension}\) | \(\ds \mathsf {M L^2 T^{-1} }\) | Definition of Planck's Constant | |||||||||||
\(\ds \E\) | \(\text {has dimension}\) | \(\ds \mathsf {I T}\) | Definition of Elementary Charge | |||||||||||
\(\ds m_\E\) | \(\text {has dimension}\) | \(\ds \mathsf M\) | Definition of Mass of Electron | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds a_0\) | \(\text {has dimension}\) | \(\ds \dfrac {\mathsf {M^{-1} L^{-3} T^4 I^2} \cdot \paren {\mathsf {M L^2 T^{-1} } }^2} {\paren {\mathsf {I T} }^2 \cdot \mathsf M}\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\mathsf {M L T^2 I^2} } {\mathsf {M I^2 T^2} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \mathsf L\) |
$\blacksquare$